"Singular homology = simplicial homology" relative to a fibration

Question

Let $p:E\to B$ be a fibration. Suppose $B$ has a simplicial decomposition. For each $n\in\mathbb{Z}_{\ge0}$, let $C_n$ be the free abelian group generated by the set of pairs $(\sigma,\tau)$ where $\sigma:\Delta^n\to E$ is a singular simplex, $\tau:\Delta^n\to B$ is a simplicial simplex, and $\tau=p\circ\sigma$. There is a boundary operator $\partial:C_n\to C_{n-1}$ defined in the usual way, making use of the faces of $\Delta^n$. Clearly $\partial\circ\partial=0$, and $(C_*,\partial)$ is a chain complex.

Update: If $B$ is a finite-dimensional simplicial complex, then for $n>\dim B$, $C_n$ is defined inductively as follows. For $n=\dim B+1$, let $C_{\dim B+1}$ be the free abelian group generated by the set of pairs $(\sigma,\tau)$ where $\sigma:\Delta^n\to E$ is a singular simplex, $\tau:\Delta^n\to B$ is a singular simplex such that $\tau=p\circ\sigma$ and $\partial(\sigma,\tau)\in C_{\dim B}$, where $\partial(\sigma,\tau)$ is defined in the usual way. Define $C_n$ similarly for all larger $n$.

My question is: Is the homology of $(C_*,\partial)$ isomorphic to the singular homology of $E$?

Answer 1

My question is: Is the homology of $(C_*,\partial)$ isomorphic to the singular homology of E?

Yes. Observe that the singular complex functor sends Serre fibrations to Kan fibrations. Thus, the map $$\def\Sing{\mathop{\rm Sing}} \Sing p\colon \Sing E\to \Sing B$$ is a Kan fibration.

Denote by $T$ the underlying simplicial set of the triangulation of $B$ so that we have a homeomorphism $|T|→B$ and an adjoint simplicial weak equivalence $t\colon T→\Sing B$.

Now the chain complex $C$ is simply the simplicial chains of the pullback of the span of simplicial sets $$T→\Sing B←\Sing E.$$

The base change of $\Sing p$ along $t$ yields a map $P→T$.

One leg of the above span is a Kan fibration, therefore the pullback computes the homotopy pullback. The map $T→\Sing B$ is a weak equivalence, hence so is its base change $P\to \Sing E$.

Thus, the map $P→T$ is weakly equivalent to $\Sing p$, so the chain complex of $P$ is quasi-isomorphic to the chain complex of $\Sing E$. The chain complex of $P$ is precisely $C$.